Let $g$ be a polynomial function and let $g'$, its derivative, be defined as $g'(x)=x^2(x-1)(x+3)$. At how many points does the graph of $g$ have a relative maximum ? Choose 1 answer: Choose 1 answer: (Choice A) A None (Choice B) B One (Choice C) C Two (Choice D) D Three
Solution: We can find the relative extrema (i.e. minima and maxima) of $g$ by looking for the intervals where its derivative $g'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. We are given that $g'(x)=x^2(x-1)(x+3)$. $g'(x)=0$ for $x=-3,0,1$. Since $g'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=-3$, $x=0$, and $x=1$. $g$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into four intervals: $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $x< \llap{-}3$ $\llap{-}3<x<0$ $0<x<1$ $x>1$ Let's evaluate $g'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g'(x)$ Verdict $x<-3$ $x=-4$ $g'(-4)=80>0$ $g$ is increasing $\nearrow$ $-3<x<0$ $x=-2$ $g'\left(-2\right)=-12<0$ $g$ is decreasing $\searrow$ $0<x<1$ $x=\dfrac12$ $g'\left(\dfrac12\right)=-\dfrac{7}{16}<0$ $g$ is decreasing $\searrow$ $x>1$ $x=2$ $g'(2)=20>0$ $g$ is increasing $\nearrow$ Now let's look at the critical points: $x$ Before After Verdict $-3$ $\nearrow$ $\searrow$ Maximum $0$ $\searrow$ $\searrow$ Not an extremum $1$ $\searrow$ $\nearrow$ Minimum Now we can see that $g$ has one relative maximum.